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How do you solve 10C3? - Answers

To solve ( 10C3 ), which represents the number of combinations of 10 items taken 3 at a time, you use the formula: [ nCr = \frac{n!}{r!(n-r)!} ] For ( 10C3 ), this becomes: [ 10C3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} ] Calculating this gives: [ = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 ] Thus, ( 10C3 = 120 ).



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How do you solve 10C3? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_10C3

To solve ( 10C3 ), which represents the number of combinations of 10 items taken 3 at a time, you use the formula: [ nCr = \frac{n!}{r!(n-r)!} ] For ( 10C3 ), this becomes: [ 10C3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} ] Calculating this gives: [ = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 ] Thus, ( 10C3 = 120 ).



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_10C3

How do you solve 10C3? - Answers

To solve ( 10C3 ), which represents the number of combinations of 10 items taken 3 at a time, you use the formula: [ nCr = \frac{n!}{r!(n-r)!} ] For ( 10C3 ), this becomes: [ 10C3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} ] Calculating this gives: [ = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 ] Thus, ( 10C3 = 120 ).

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      To solve ( 10C3 ), which represents the number of combinations of 10 items taken 3 at a time, you use the formula: [ nCr = \frac{n!}{r!(n-r)!} ] For ( 10C3 ), this becomes: [ 10C3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} ] Calculating this gives: [ = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 ] Thus, ( 10C3 = 120 ).

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